Question

graph the equation $y = -x^2 + 2x + 3$ on the accompanying set of axes. you must plot 5 points including the roots and the vertex. using the graph, determine the vertex of the parabola.
click to plot points. click points to delete them.

Explanation:

Step1: Find the roots (x-intercepts)

Set \( y = 0 \), so \( -x^2 + 2x + 3 = 0 \). Multiply both sides by -1: \( x^2 - 2x - 3 = 0 \). Factor: \( (x - 3)(x + 1) = 0 \). Thus, roots are \( x = 3 \) and \( x = -1 \). So points are \( (-1, 0) \) and \( (3, 0) \).

Step2: Find the vertex

For quadratic \( y = ax^2 + bx + c \), x-coordinate of vertex is \( x = -\frac{b}{2a} \). Here, \( a = -1 \), \( b = 2 \). So \( x = -\frac{2}{2(-1)} = 1 \). Substitute \( x = 1 \) into equation: \( y = -(1)^2 + 2(1) + 3 = -1 + 2 + 3 = 4 \). So vertex is \( (1, 4) \).

Step3: Find two more points

Let's pick \( x = 0 \): \( y = -0 + 0 + 3 = 3 \), so point \( (0, 3) \).
Pick \( x = 2 \): \( y = -(4) + 4 + 3 = 3 \), so point \( (2, 3) \).

Now we have 5 points: \( (-1, 0) \), \( (3, 0) \), \( (1, 4) \), \( (0, 3) \), \( (2, 3) \). Plotting these, the vertex is \( (1, 4) \).

Answer:

The vertex of the parabola is \((1, 4)\)